When `1xx10^(-3)` " mol of "the chloride of an elements Y was completely hydrolysed, it was found that the resulting solution requried 20 " mL of " 0.1 M aqueous silver nitrate for complete precipitation of the chloride ion. Elements Y could be

Options

A. Aluminium
B. Phosphorus
C. Silicon
D. Sulphur

Explanation

To solve the problem step by step, we will analyze the information given and derive the necessary conclusions.

### Step 1: Understand the Problem
We are given that 1 x 10^(-3) moles of a chloride of element Y was hydrolyzed, and the resulting solution required 20 mL of 0.1 M silver nitrate (AgNO3) for complete precipitation of chloride ions.

### Step 2: Calculate Moles of AgNO3 Used
First, we need to calculate the number of moles of AgNO3 used in the reaction. The formula to calculate moles is given by:

\[ \text{Moles} = \text{Molarity} \times \text{Volume (in L)} \]

Converting 20 mL to liters:

\[ 20 \text{ mL} = 20 \times 10^{-3} \text{ L} = 0.020 \text{ L} \]

Now, substituting the values into the formula:

\[ \text{Moles of AgNO3} = 0.1 \text{ M} \times 0.020 \text{ L} = 0.002 \text{ moles} \]

### Step 3: Relate Moles of AgNO3 to Moles of Chloride Ion
From the reaction, we know that 1 mole of AgNO3 reacts with 1 mole of chloride ions (Cl^-). Therefore, the moles of chloride ions produced from the hydrolysis of YCl_n will also be 0.002 moles.

### Step 4: Relate Moles of YCl_n to Moles of Chloride Ion
The problem states that 1 x 10^(-3) moles of YCl_n produces chloride ions. If we denote the number of chloride ions produced from YCl_n as n, we have:

\[ 1 \times 10^{-3} \text{ moles of YCl}_n = n \text{ moles of Cl}^- \]

Since we found that the moles of Cl^- is 0.002 moles, we can set up the equation:

\[ 1 \times 10^{-3} = n \]

### Step 5: Solve for n
From the previous equation, we can see that:

\[ n = 0.002 \text{ moles} \]

### Step 6: Determine the Formula of the Chloride
Since we have determined that n = 2, this implies that the chloride of element Y must be YCl2.

### Step 7: Identify Element Y
Now we need to identify which element Y can form a chloride with the formula YCl2. We can analyze the options provided:

- Aluminum (Al) forms AlCl3.
- Phosphorus (P) forms PCl3 and PCl5.
- Silicon (Si) forms SiCl4.
- Sulfur (S) forms SCl2.

The only element that forms a chloride with the formula YCl2 is sulfur (S).

### Conclusion
Thus, the element Y is **sulfur**.

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